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In proving closed algebraic set in $\mathbb P_n(k)$ has decomposition into irreducible algebraic sets, we can argue by decomposition in noetherian space.

Can we prove directly by decomposition of homogeneous ideal into homogeneous prime ideals?

I don't know if this holds. I tried to use the correspondence between ideals in $k[X_1,...,X_n]$ and homogeneous ideals in $k[X_0,...,X_n]$ by $I \rightarrow {}^hI$ and ${}^aJ \leftarrow J$. Suppose we have $J$ a homogeneous ideal in $k[X_0,...,X_n]$, then ${}^aJ = \mathfrak{q}_1\cap\cdots\cap\mathfrak{q}_k$ and we have ${}^h({}^aJ) = {}^h\mathfrak{q}_1\cap\cdots\cap{}^h\mathfrak{q}_k$ a primary decomposition. But ${}^h({}^aJ)=J$ may not hold. Take $J=(X_0,...,X_n)$ for example: ${}^h({}^aJ)=k[X_0,..,X_n]$

I saw tags here: "minimal primes of a homogeneous ideal are homogeneous" but the problem was not fully answered and the cited link is missing.

Can anyone answer that tag or help me with this question?

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1 Answers1

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It's true that the homogeneous ideals have homogeneous primary decompositions in any finitely generated graded algebra over a field. This follows from the usual primary decomposition by taking the homogeneous parts of primary ideals involved into decomposition. (For more details see here, Section 1.4.)

For the part of your question about the minimal primes of homogeneous ideals look here.

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    May I ask another question? Is there any way to tell when do ${}^h{}^aJ$ and $J$ coincide? Thanks! –  Oct 14 '13 at 13:50
  • @mqx Maybe you should post this as a separate question. The algebraic geometers can answer you better than me. –  Oct 14 '13 at 14:00